Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
const int inf = 1e9;

vector<int> bfs(vector<vector<int> > &adj, int s){
	vector<int> ret(adj.size(), inf);
	ret[s] = 0;
	queue<pair<int,int> > wait;
	wait.push(make_pair(s, 0));
	while(!wait.empty()){
		int pos = wait.front().first;
		int cost = wait.front().second;
		wait.pop();
		for(int i=0; i<(int)adj[pos].size(); i++){
			if(ret[adj[pos][i]] == inf){
				ret[adj[pos][i]] = cost+1;
				wait.push(make_pair(adj[pos][i], cost+1));
			}
		}
	}
	return ret;
}

int main(){
	int n,m,s,t;
	cin >> n >> m >> s >> t;
	vector<vector<int> > adj(n+1);
	vector<int> x(m),y(m);
	for(int i=0; i<m; i++){
		cin >> x[i] >> y[i];
		adj[x[i]].push_back(y[i]);
		adj[y[i]].push_back(x[i]);
	}
	
	vector<int> froms = bfs(adj, s);
	vector<int> fromt = bfs(adj, t);
	int dist_st = froms[t];
	vector<long long int> bsums(n+1, 0), bsumt(n+1, 0);
	for(int i=1; i<n+1; i++){
		if(froms[i]!=inf) bsums[froms[i]]++;
		if(fromt[i]!=inf) bsumt[fromt[i]]++;
	}
	
	long long int ans = 0;
	for(int i=0; i<dist_st -1; i++){
		int j=dist_st -i -2;
		if(i<j){
			ans += bsums[i]*bsumt[j] +bsums[j]*bsumt[i];
		}else if(i==j){
			ans += bsums[i]*bsumt[j];
		}
	}
	for(int i=0; i<m; i++){
		if(froms[x[i]] + fromt[y[i]] == dist_st-2 ||
		froms[y[i]] + fromt[x[i]] == dist_st-2){
			 ans--;
		}
	}
	cout << ans << endl;
	
	return 0;
}